How do you find all zeros with multiplicities of $f (x) = 2 x^{4} - 7 x^{2} + 6$ $f(x)=2x^4-7x^2+6$ ?

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Answer 1 · 2017-05-28T21:09:00

See explanation.

The original equation can be transformed to a quadratic equation by a substitution: $x^{2} = t$ $x^2=t$

$2 t^{2} - 7 t + 6 = 0$ $2t^2-7t+6=0$

$Delta=(-7)^2-4*2*6=49-48=1$

$sqrt(Delta)=1$

$t_1=(7-1)/4=3/2$ and $t_2=(7+1)/4=2$

Now we have to find corresponding $x$ values:

$x^2=3/2$

$x_1=-sqrt(3/2)$ and $x_2=sqrt(3/2)$

If we rationalize the denominators we get:

$x_1=-sqrt(6)/2$ and $x_2=sqrt(6)/2$

$x^2=2$

$x_3=-sqrt(2)$ and $x_4=-sqrt(2)$

Final answer:

The function has 4 real zeros:

$x_1=-sqrt(6)/2$ , $x_2=sqrt(6)/2$ , $x_3=-sqrt(2)$ , $x_4=-sqrt(2)$

How do you find all zeros with multiplicities of f(x)=2x^4-7x^2+6f(x)=2x4−7x2+6f(x)=2x^4-7x^2+6?